Problem: In Takahashi's mind, there is always an integer sequence of length 2 \times 10^9 + 1: A = (A_{-10^9}, A_{-10^9 + 1}, ..., A_{10^9 - 1}, A_{10^9}) and an integer P.

Initially, all the elements in the sequence A in Takahashi's mind are 0, and the value of the integer P is 0.

When Takahashi eats symbols `+`, `-`, `>` and `<`, the sequence A and the integer P will change as follows:

* When he eats `+`, the value of A_P increases by 1;
* When he eats `-`, the value of A_P decreases by 1;
* When he eats `>`, the value of P increases by 1;
* When he eats `<`, the value of P decreases by 1.



Takahashi has a string S of length N. Each character in S is one of the symbols `+`, `-`, `>` and `<`. He chose a pair of integers (i, j) such that 1 \leq i \leq j \leq N and ate the symbols that are the i-th, (i+1)-th, ..., j-th characters in S, in this order. We heard that, after he finished eating, the sequence A became the same as if he had eaten all the symbols in S from first to last. How many such possible pairs (i, j) are there?

Constraints

* 1 \leq N \leq 250000
* |S| = N
* Each character in S is `+`, `-`, `>` or `<`.

Input

Input is given from Standard Input in the following format:


N
S


Output

Print the answer.

Examples

Input

5
+>+<-


Output

3


Input

5
+>+-<


Output

5


Input

48
-+><<><><><>>>+-<<>->>><<><<-+<>><+<<>+><-+->><<


Output

475

Solution (in CPP):

Solution:
#include <iostream>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <deque>
#include <queue>
#include <vector>
#include <map>
using namespace std;
map<unsigned long long,int>M;
const unsigned long long key=1e9+7,N=260000;
unsigned long long invkey;
char ch[N+10];
int n;
unsigned long long quick(unsigned long long k1,unsigned long long k2){
	unsigned long long k3=1;
	while (k2){
		if (k2&1) k3*=k1;
		k1*=k1; k2>>=1;
	}
	return k3;
}
unsigned long long mul=0,add=0;
unsigned long long getall(){
	unsigned long long k1=1,ans=0;
	for (int i=1;i<=n;i++)
		if (ch[i]=='+') ans+=k1;
		else if (ch[i]=='-') ans-=k1;
		else if (ch[i]=='>') k1*=key;
		else k1*=invkey;
	return ans;
}
unsigned long long getb(unsigned long long ke,unsigned long long a,unsigned long long b){
	return (ke-b)*quick(a,(1ull<<63)-1);
}
int main(){
	scanf("%d",&n);
	scanf("%s",ch+1);
	invkey=quick(key,(1ull<<63)-1);
	unsigned long long ke=getall();
	unsigned long long a=1,b=0;
	long long ans=0;
	for (int i=n;i;i--){
		ans+=M[getb(ke,a,b)];
		M[getb(0,a,b)]++;
		if (ch[i]=='+') b++;
		else if (ch[i]=='-') b--;
		else if (ch[i]=='>') b*=key,a*=key;
		else if (ch[i]=='<') b*=invkey,a*=invkey;
	}
	ans+=M[getb(ke,a,b)];
	cout<<ans<<endl;
}